Omar decided to paint some of the rooms at his 28-room inn, Omar's Place. He discovered he needed $\frac{3}{5}$ of a can of paint per room. If Omar had 9 cans of paint, how many rooms could he paint?
Solution: We can divide the cans of paint (9) by the paint needed per room ( $\frac{3}{5}$ of a can) to find out how many rooms Omar could paint. $ \dfrac{{9 \text{ cans of paint}}} {{\dfrac{3}{5} \text{ can per room}}} = {\text{ rooms}} $ Dividing by a fraction is the same as multiplying by the reciprocal. The reciprocal of ${\dfrac{3}{5} \text{ can per room}}$ is ${\dfrac{5}{3} \text{ rooms per can}}$ $ {9\text{ cans of paint}} \times {\dfrac{5}{3} \text{ rooms per can}} = {\text{ rooms}} $ ${\dfrac{45}{3}\text{ rooms}} = 15\text{ rooms}$ Omar could paint 15 rooms.